The contour integrals I am familiar with end up on the same branch of the function as they started, perhaps circulating a pole. It is not clear whether integrating √z just once around the origin makes sense. Lets try.
For practice we integrate z around the unit circle. I won’t indulge in mathematical notation to depict the path but assume that we integrate about the unit circle, beginning and ending at 1. This is the limit of summing z dz but when z is near one then dz is in the direction of i! The indefinite integral, p, from 1 to x, as x moves along the unit circle begins to move up from the origin as x moves away from 1. When x reaches i the motion of p is towards −i. By the time x gets to −1 p has moved completely around a circle thru the origin and centered at −½. When x gets back to 1 p has circulated its circle twice. (If we had integrated a constant then p’s direction would have gone thru one circle and p would have ended at 0.) Bizzare! The integral is 0 as we have been taught to believe. If we had integrated 1/z then p would have moved up at a constant velocity and ended at 2πi, also as we have been led to believe.
Now we go for √z around the same path. Again p is the indefinite integral. p begins moving as it did before but its direction of motion turns slower. In both cases its speed is a constant 1. When z gets to i, p is moving in the direction e5πi/4. When z gets back to 1 the direction of p has changed thru 3π and we have traversed half a circle thru the origin. As z returns to 1 the velocity of p is −i and p is not 0! Remember √z is on another branch with the opposite sign. If we integrate twice around, p returns to 0.
This does not bode well for the dilogarithm for the integrand does not repeat as z circulates, it continues to drift in the general direction i as we spiral the infinite set of branches of the logarithm.
This program numerically computes the integral around the branch point 1t 1 of −log(1−z)/z in order to evaluate the dilogarithm on different branches. The routine ln there is the complex logarithm but remembers which branch you are on and demands that you move slowly. We integrate counterclockwise around the unit circle with center = 1, beginning and ending at 0.
Oops, debugging this program brought me to realize that there are poles(!) on the other branches of the integrand. Despite appearances −log(1−z)/z behaves nicely at 0, but only as long as we are on the branch of the log that goes to 0 there. The only such branch is the ‘main branch’. The other branches of dilogarithm have a pole at 0.